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I am interested in simplifying the integral $$\int \frac{1}{x^2}\exp(-ax+b-\frac{c}{x})dx$$ with $a, b, c \in \mathbb{R}$. Do you have any idea? With $a=0$ or $c=0$ I know the solutions but what about the general case $a \neq 0, c \neq 0$?

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    $\begingroup$ The u-substitution u = -1/x jumps out at me. $\endgroup$ – Kaynex Nov 5 '16 at 14:36
  • $\begingroup$ Wolfram alpha says there's no closed form using elementary functions. $\endgroup$ – Sophie Nov 5 '16 at 14:56
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    $\begingroup$ @Sophie Wolfram Alpha is not as smart as some of us. $\endgroup$ – Turing Nov 5 '16 at 15:07
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    $\begingroup$ @AlanTuring. As a comment to another remark of the same kind, I wrote "Well.. nobody's perfect !" to remember the movie but thinking about a few other things. $\endgroup$ – Claude Leibovici Nov 5 '16 at 15:38
  • $\begingroup$ After the substitution $x=\frac{1}{t}$, in some cases a closed form for the integral over some intervals can be found through Glasser's master theorem: mathworld.wolfram.com/GlassersMasterTheorem.html $\endgroup$ – Jack D'Aurizio Nov 5 '16 at 19:49
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We can use the substitution Kaynex suggested you:

$$u = \frac{1}{x} ~~~~~~~ \text{d}u = -\frac{1}{x^2}\ \text{d}x$$

to obtain the "easier" integral:

$$-\int e^{-a/u + b - cu}\ \text{d}u$$

Which I will rewrite it by calling $u = x$ again because I don't like the letter $u$.

$$-\int e^{-a/x + b - cx}\ \text{d}x$$

First of all we notice that we can take a term out:

$$-e^b\int e^{-a/x - cx}\ \text{d}x$$

Now the interesting part: a general solution, namely a closed form one, does not exist. First of all because we don't know the range of integration, hence the integral may exist or it mayn't. But even in the general form, it's not a closed form integral. We may think about a series expansion of the exponential but not the whole exponential. This is because we are reasoning in the general way, so we will split into two different solutions, according to what part of the exp we expand in series.

First Solution: Series of $e^{-a/x}$

Knowing that

$$e^{-a/x} = \sum_{k = 0}^{+\infty} \frac{(-a/x)^k}{k!}$$

We get the integral

$$-e^b\sum_{k = 0}^{+\infty} \frac{(-a)^k}{k!}\int \frac{e^{-cx}}{x^k}\ \text{d}x$$

This is a well known integral and its result is in terms of the incomplete Gamma function:

$$\int \frac{e^{-cx}}{x^k}\ \text{d}x = -x^{1-k} (c x)^{k-1} \Gamma (1-k,c x)$$

Hence in the first case we get: the solution

$$\boxed{e^b\sum_{k = 0}^{+\infty} \frac{(-a)^k}{k!}\left(x^{1-k} (c x)^{k-1} \Gamma (1-k,c x)\right)}$$

Second Case: Series of $e^{-cx}$

In this case

$$e^{-cx} = \sum_{k = 0}^{+\infty} \frac{(-cx)^k}{k!}$$

hence

$$e^b \sum_{k = 0}^{+\infty} \frac{(-c)^k}{k!}\int x^k e^{-a/x}\ \text{d}x$$

In this case we have another well known integral, that is

$$\int x^k e^{-a/x}\ \text{d}x = x^{k+1} E_{k+2}\left(\frac{a}{x}\right)$$

Where the special function in this case is the Exponential Integral $n$: http://mathworld.wolfram.com/En-Function.html

The solution here is

$$\boxed{-e^b\sum_{k = 0}^{+\infty} \frac{(-c)^k}{k!}x^{k+1} E_{k+2}\left(\frac{a}{x}\right)}$$

Third case, for the sake of generality: Series of the whole exp

$$e^{-a/x - cx} = \sum_{k = 0}^{+\infty} \frac{(-a/x - cx)^k}{k!}$$

Hence

$$e^b \sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!}\int (a/x + cx)^k\ \text{d}x$$

The integral is not an easy one but it does exist:

$$\int (a/x + cx)^k\ \text{d}x = -\frac{x \left(\frac{a}{x}+c x\right)^k \left(\frac{c x^2}{a}+1\right)^{-k} \, _2F_1\left(\frac{1-k}{2},-k;\frac{1-k}{2}+1;-\frac{c x^2}{a}\right)}{k-1}$$

It's expressed in terms of HyperGeometric Functions: http://mathworld.wolfram.com/HypergeometricFunction.html and https://en.wikipedia.org/wiki/Hypergeometric_function

Hence the third case gives:

$$\boxed{e^b\sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!}\frac{x \left(\frac{a}{x}+c x\right)^k \left(\frac{c x^2}{a}+1\right)^{-k} \, _2F_1\left(\frac{1-k}{2},-k;\frac{1-k}{2}+1;-\frac{c x^2}{a}\right)}{k-1}}$$

Conclusion

You may have some fun in taking the first 2-3 terms of each series, to have a quite more normal idea about the behaviors of the series. Note that there is no close forme for the firs series. In any case, with a pair of extrema in the integration like would be easier!

Happy integrating!

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  • $\begingroup$ Thanks a lot, Alan Turing! That's really a very helpful and detailed answer! In case of taking the integral from $0$ to $\infty$ I think a solution is given here math.stackexchange.com/questions/664298/… (which then has to be multiplied by $-\exp(b)$) $\endgroup$ – Heiting Nov 5 '16 at 15:38
  • $\begingroup$ @Heiting My pleasure I helped a bit! ^^ Feel free to have fun in exploring the special functions and the solutions. There are lots of amusing things to do here! :D $\endgroup$ – Turing Nov 5 '16 at 15:39
  • $\begingroup$ @Heiting Thank you so much for the link!! $\endgroup$ – Turing Nov 5 '16 at 16:08
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$\int\dfrac{1}{x^2}e^{-ax+b-\frac{c}{x}}~dx$

$=-e^b\int_x^\infty\dfrac{e^{-at-\frac{c}{t}}}{t^2}~dt+C$

$=-e^b\int_1^\infty\dfrac{e^{-axt-\frac{c}{xt}}}{(xt)^2}~d(xt)+C$

$=-\dfrac{e^b}{x}\int_1^\infty\dfrac{e^{-axt-\frac{c}{xt}}}{t^2}~dt+C$

$=-\dfrac{e^b}{x}K_1\left(ax,\dfrac{c}{x}\right)+C$ (according to https://core.ac.uk/download/pdf/81935301.pdf)

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